Affine transformation that sends a conic to itself but does not preserve the focci or the axes

Question

So I'm trying to find an affine transformation that sends a conic to itself but does not preserve the foci or the axes. I don't know if this can be done. I'm pretty sure that if it is possible then I have to do a sort of "conjugation" of transformations. But nothing I've tried seems to work.

This is sort of what I mean http://www.maths.gla.ac.uk/wws/cabripages/klein/affinesymmetry.html

Thanks for any help.

Answer 1

Let $\lambda>0$ and $\lambda\neq 1$. The transformation $$(x,y)\mapsto(\lambda x, \lambda^{-1} y)$$ maps the hyperbola $xy=1$ to itself. It does not fix the foci or the symmetry axes. For any angle $\varphi$ that is not a multiple of $\pi$ the transformation $$(x, y)\mapsto(\cos(\varphi) x -\lambda \sin(\varphi) y, \lambda^{-1} \sin(\varphi) x+\cos(\varphi)y)$$ maps the ellipse $x^2+\lambda^2 y^2=1$ to itself but again it does not fix its foci or symmetry axes. The transformation $$(x,y)\mapsto (x+\lambda, y+2\lambda x+\lambda^2)$$ maps the parabola $y=x^2$ to itself but...

Answer 2

If a conic is given by the points $(x,y)\in\mathbb{R}^2$ such that: $$ ax^2+bxy+cy^2+dx+ey+f = 0 \tag{1}$$ then the foci and axis of the conic depend on $(a,b,c,d,e,f)\in\mathbb{P}^5(\mathbb{R})$. If a conic is mapped to itself by an affine map, then its equation $(1)$ stays the same, so the foci and axis stay the same, too.

If that wasn't convincing enough, again: if there is an ellipse/parabola/hyperbola on the euclidean plane, you can find its foci and axis by straightedge and compass. So same conic, same foci and axis.